The daily water consumption for an Ohio community is normally distributed with a mean consumption of 750,000 gallons and a standard deviation of 80,000 gallons. The community water system will experience a noticeable drop in water pressure when the daily water consumption exceeds 984,000 gallons. What is the probability of experiencing such a drop in water pressure?

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The speed of limit on a particular freewayis 65mph. A radar system revealed that the actual speed of traffic is approximately normally dostributed with the average of 68mph and standard deviation of 4mph.

What proportion of cars exceed the speed limit

To find the probability of experiencing a drop in water pressure when the daily water consumption exceeds 984,000 gallons, we need to calculate the z-score and then find the corresponding probability.

The z-score represents the number of standard deviations a value is from the mean and is calculated using the formula:

z = (x - μ) / σ

Where:
- x is the value we want to find the probability for (984,000 gallons in this case)
- μ is the mean (750,000 gallons)
- σ is the standard deviation (80,000 gallons)

Substituting the values into the formula, we get:

z = (984,000 - 750,000) / 80,000

Simplifying the equation, we have:

z = 234,000 / 80,000
z = 2.925

Now, with the z-score calculated, we can use a standard normal distribution table or a calculator to find the corresponding probability.

Looking up the z-score of 2.925 in a standard normal distribution table, we find that the probability is approximately 0.9981.

Therefore, the probability of experiencing a drop in water pressure when the daily water consumption exceeds 984,000 gallons is approximately 0.9981, or 99.81%.